Math Problem Statement
Use the graph to determine the values of a for which
lim x→a f(x)
exists. (Enter your answer using interval notation.)
Solution
To determine the values of for which exists, we need to analyze the continuity and behavior of the piecewise function at each segment and at the boundaries where the function definition changes.
Function Recap:
\sqrt[3]{x} & \text{if } x \leq -1 \\ x & \text{if } -1 < x \leq 2 \\ (x - 1)^2 & \text{if } x > 2 \end{cases}$$ ### Analysis of Continuity: 1. **For $$x \leq -1$$**: - The function $$f(x) = \sqrt[3]{x}$$ is continuous for all $$x$$ in this interval. - Therefore, the limit $$\lim_{x \to a} f(x)$$ exists for all $$a \leq -1$$. 2. **For $$-1 < x \leq 2$$**: - The function $$f(x) = x$$ is continuous for all $$x$$ in this interval. - Therefore, the limit $$\lim_{x \to a} f(x)$$ exists for all $$a$$ in the interval $$(-1, 2]$$. 3. **For $$x > 2$$**: - The function $$f(x) = (x - 1)^2$$ is a quadratic function, which is continuous for all $$x > 2$$. - Therefore, the limit $$\lim_{x \to a} f(x)$$ exists for all $$a > 2$$. 4. **At the boundaries $$x = -1$$ and $$x = 2$$**: - **At $$x = -1$$**: - From the left, $$\lim_{x \to -1^-} f(x) = \sqrt[3]{-1} = -1$$. - From the right, $$\lim_{x \to -1^+} f(x) = -1$$. - Since both limits match and equal the function value $$f(-1) = -1$$, the limit exists at $$x = -1$$. - **At $$x = 2$$**: - From the left, $$\lim_{x \to 2^-} f(x) = 2$$. - From the right, $$\lim_{x \to 2^+} f(x) = (2 - 1)^2 = 1$$. - Since the left-hand limit does not equal the right-hand limit, the limit $$\lim_{x \to 2} f(x)$$ does **not** exist at $$x = 2$$. ### Conclusion: The limit $$\lim_{x \to a} f(x)$$ exists for the following values of $$a$$: - $$a \leq -1$$ - $$-1 < a < 2$$ - $$a > 2$$ This can be written in interval notation as: $$(-\infty, 2) \cup (2, \infty)$$ Would you like to discuss this further or see a visual explanation? ### Follow-up Questions: 1. Why does the limit not exist at $$x = 2$$? 2. How would the interval change if the function was defined differently at $$x = 2$$? 3. What is the importance of continuity in determining the existence of limits? 4. How can one determine whether a function is continuous at a point? 5. What is the difference between one-sided limits and two-sided limits? **Tip:** When analyzing the existence of a limit, always consider the behavior of the function from both sides of the point of interest. If the left-hand limit equals the right-hand limit, the limit exists at that point.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Continuity
Piecewise Functions
Formulas
-
Theorems
-
Suitable Grade Level
Grades 11-12
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